Question

The numbers from $ 1 $ to $ 1996 $ are written down:
 $ 12345678910111213.........19951996 $
How many zeroes are there?

Answer 1

Hint: The given question is related to the number system topic. The question asks us to find the number of zeroes that will be present in a series of the numbers written from $ 1 $ to $ 1996 $ side by side. The question seems quite daunting at first because we cannot calculate this question manually in a reasonable amount of time, so we will try to see the pattern of the series, break them into various little series with the same number of zeroes and get our answer after adding all of them.

Complete step-by-step answer:
We have to calculate the zeros of a number present in the numbers from $ 1 $ to $ 1996 $ when written side by side.
For two digit numbers:
From $ 0 $ to $ 99 $ there are $ 9 $ zeroes at,
 $ 10,20,30,40............. $
Total: $ 9 $
For three digit numbers:
From $ 100 $ to $ 999 $ , there will be:
 $ 18 $ zeros for each interval from $ 101 - 199,201 - 299 $ $ ...... $
Total Zeroes: $ 9 \times 18 = 162 $
 $ 2 $ zeroes for each of $ 100,200,300........ $
 $ \Rightarrow 2 \times 9 = 18 $
Hence total zeroes of three digit numbers are :
 $ 18 + 162 = 180 $
For four digit numbers:
From $ 1001 - 1099 $
there will be $ 99 + 9 + 9 = 117 $ ( $ 99 $ hundred place zeroes, $ 9 $ tens place zeroes and $ 9 $ one’s place zeroes)
Similarly for $ 1101 - 1199,............1801,1899 $
there will be $ 8 \times (9 + 9) = 144 $ zeroes (as there are $ 8 $ interval of $ 9 $ ones and $ 9 $ tens
For $ 1901 - 1996 $
there will be $ 18 $ zeroes
For the term $ 1000 $ there will be $ 3 $ zeroes
And for terms like $ (1100,1200,1300,.......1900) $
there will be $ 9 \times 2 = 18 $ zeroes.
Total for four digits: $ 117 + 144 + 18 + 18 + 3 = 300 $
Grand total of all the numbers: $ 300 + 180 + 9 = 489 $ zeroes.
So, the correct answer is “ 489 ZEROS”.

Note: We can also do this question by using permutations and combinations. Each type of digits i.e. two, three, and four digits numbers can each be individually solved for permutation of zeroes at places except for the highest place of the number i.e. tens in case of two digit number, hundred place in case of three digit number and so on. The permutation should be applied individually at each place value rather than using the permutation formula.